Question

3.5cm needle is placed 10cm away from a convex mirror of focal length 15cm.Give the location of image and magnification. Describe what happens to the image as the needle is moved farther from the mirror

Answer 1

Given that,

Height of needle = 3.5 cm

Distance = 10 cm

Focal length= 15 cm

We need to calculate the image distance

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

$\dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{10}$

$\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{10}$

$\dfrac{1}{v}=\dfrac{1}{6}$

$v=6\ cm$

We need to calculate the height of the image

Using formula of magnification

$m=\dfrac{h'}{h}=\dfrac{-v}{u}$

$h'=\dfrac{-v}{u}\times h$

Put the value into the formula

$h'=\dfrac{-6}{-10}\times3.5$

$h'=2.1\ cm$

The height of the image is 2.1cm.

So, The image is erect, virtual, and diminished.

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{-v}{u}$

Put the value into the formula

$m=\dfrac{-6}{-10}$

$m=0.6$

We need to find what happens to the image as the needle is moved farther from the mirror

According to data,

When the needle is moved farther from the mirror then the size of the image reduce slowly slowly.

Hence, The image of the needle is 6 cm away from the mirror and it is on the other side of the mirror.

If the needle is moved farther from the mirror then the size of the image will reduce slowly slowly.